TITLE: Study the convergence of $x_n$, $ x_{n+1}=\frac{1}{2}\big(\frac{x_n+3}{ x_n}\big)$, with $x_0=1.$
QUESTION [5 upvotes]: How can we prove that the limit of the sequence $x_n$ defined by:
$ x_{n+1}=\frac{1}{2}\big(\frac{x_n+3}{ x_n}\big)$, with $x_0=1$ exists? I have tried to prove that it is Cauchy , but I failed? Can I get some help, and thanks in advance.

REPLY [2 votes]: Just rewriting$$x_{n+1}=\frac{1}{2}\left(\frac{x_n+3}{ x_n}\right)\implies x_{n+1}=x_n-\frac{2 x_n^2-x_n-3}{2 x_n}$$ and recognize the Newton formula for finding the zero's of function $$f(x)=2x^2-x-3$$